The equation of a circle $C$ is $x^2+y^2-12x-13 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-12x) + (y^2) = 13$ $(x^2-12x+36) + (y^2) = 13 + 36 + 0$ $(x-6)^{2} + y^2 = 49 = 7^2$ Thus, $(h, k) = (6, 0)$ and $r = 7$.